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A 50.0 mL of 0.88 M H2O2 and 10.0 mL of 0.50 M Fe(NO3)3 were combined...

A 50.0 mL of 0.88 M H2O2 and 10.0 mL of 0.50 M Fe(NO3)3 were combined and a temperature change of 6.61 was observed. The specific heat of water is 4.18 J/(g * ∘C)

Calculate the total heat change (in kilojoules) for the calorimeter. Record your answer with the proper significant figures and include the correct sign if needed.


Assume the density and specific heat of the solution are the same as that of water.

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